question_answer

Three identical thin rods, each of mass \[m\] and length \[\ell \], are joined to form an equilateral triangular frame. The moment of inertia of the frame about an axis parallel to its one side and passing through the opposite vertex is

A) \[\frac{5}{2}m{{\ell }^{2}}\]  

B) \[\frac{5}{4}m{{\ell }^{2}}\]

C) \[\frac{3}{2}m{{\ell }^{2}}\]  

D) \[\frac{5}{3}m{{\ell }^{2}}\]

Correct Answer: B

Solution :

[b] Moment of Inertia of each of rod AC and BC about the given axis OO' is \[{{I}_{AC}}={{I}_{BC}}=\frac{m{{\ell }^{2}}}{3}{{\sin }^{2}}60{}^\circ =\frac{m{{\ell }^{2}}}{4}\] and M.I. of rod AB about the given axis OO' is \[{{I}_{AB}}=m{{\left( \frac{\ell \sqrt{3}}{2} \right)}^{2}}=\frac{3}{4}m{{\ell }^{2}}\] Hence, \[I={{I}_{AC}}+{{I}_{BC}}+{{I}_{AB}}=\frac{m{{\ell }^{2}}}{4}+\frac{m{{\ell }^{2}}}{4}+\frac{3}{4}m{{\ell }^{2}}=\frac{5}{4}m{{\ell }^{2}}\]